Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, u, v, eNum;
const int maxn = 100005;
struct node {
  int x, y, next;
} a[2 * maxn];
int head[maxn], depth[maxn];
int anc[maxn][18];
int minNode[maxn][18][12];
int man[maxn][12];
int aa;
void add(int u, int v) {
  eNum++;
  a[eNum].x = u;
  a[eNum].y = v;
  a[eNum].next = head[u];
  head[u] = eNum;
  eNum++;
  a[eNum].x = v;
  a[eNum].y = u;
  a[eNum].next = head[v];
  head[v] = eNum;
}
void merge(int destt[], int arr1[], int arr2[]) {
  int i1 = 1;
  int i2 = 1;
  int dest[11];
  dest[0] = 0;
  while (dest[0] < 10 && i1 <= arr1[0] && i2 <= arr2[0]) {
    if (arr1[i1] < arr2[i2]) {
      dest[0]++;
      dest[dest[0]] = arr1[i1];
      i1++;
    } else {
      if (arr1[i1] > arr2[i2]) {
        dest[0]++;
        dest[dest[0]] = arr2[i2];
        i2++;
      } else {
        dest[0]++;
        dest[dest[0]] = arr2[i2];
        i1++;
        i2++;
      }
    }
  }
  if (dest[0] < 10) {
    if (i1 <= arr1[0]) {
      while (i1 <= arr1[0] && dest[0] < 10) {
        dest[0]++;
        dest[dest[0]] = arr1[i1];
        i1++;
      }
    }
    if (i2 <= arr2[0]) {
      while (i2 <= arr2[0] && dest[0] < 10) {
        dest[0]++;
        dest[dest[0]] = arr2[i2];
        i2++;
      }
    }
  }
  for (int i = 0; i <= dest[0]; i++) destt[i] = dest[i];
}
void dfs(int now, int pre, int dep) {
  anc[now][0] = pre;
  minNode[now][0][0] = man[pre][0];
  depth[now] = dep;
  for (int i = 1; i <= man[pre][0]; i++) {
    minNode[now][0][i] = man[pre][i];
  }
  int i = head[now];
  while (i != 0) {
    if (a[i].y != pre) {
      dfs(a[i].y, now, dep + 1);
    }
    i = a[i].next;
  }
}
void initialize() {
  bool flag;
  for (int i = 1; i <= 17; i++) {
    flag = true;
    for (int j = 1; j <= maxn; j++) {
      if ((depth[j] - (1 << i)) > 0) {
        flag = false;
        anc[j][i] = anc[anc[j][i - 1]][i - 1];
        merge(minNode[j][i], minNode[j][i - 1], minNode[anc[j][i - 1]][i - 1]);
      }
    }
    if (flag) break;
  }
}
void lca(int u, int v, int aa) {
  if (depth[u] > depth[v]) swap(u, v);
  int delta = depth[v] - depth[u];
  int vv = v;
  int uu = u;
  int ans[12];
  for (int i = 0; i <= man[v][0]; i++) ans[i] = man[v][i];
  for (int i = 0; (1 << i) <= delta; i++) {
    if ((1 << i) & delta) {
      v = anc[v][i];
    }
  }
  if (v == u) {
    delta = depth[vv] - depth[v];
    for (int i = 0; (1 << i) <= delta; i++) {
      if ((1 << i) & delta) {
        merge(ans, ans, minNode[vv][i]);
        vv = anc[vv][i];
      }
    }
    printf("%d", min(ans[0], aa));
    for (int i = 1; i <= min(ans[0], aa); i++) printf(" %d", ans[i]);
    printf("\n");
    return;
  }
  for (int i = 17; i >= 0; i--) {
    if (anc[v][i] != anc[u][i]) {
      v = anc[v][i];
      u = anc[u][i];
    }
  }
  int lcaN = anc[u][0];
  delta = depth[vv] - depth[lcaN];
  for (int i = 0; (1 << i) <= delta; i++) {
    if ((1 << i) & delta) {
      merge(ans, ans, minNode[vv][i]);
      vv = anc[vv][i];
    }
  }
  merge(ans, ans, man[uu]);
  delta = depth[uu] - depth[lcaN];
  for (int i = 0; (1 << i) <= delta; i++) {
    if ((1 << i) & delta) {
      merge(ans, ans, minNode[uu][i]);
      uu = anc[uu][i];
    }
  }
  printf("%d", min(ans[0], aa));
  for (int i = 1; i <= min(ans[0], aa); i++) printf(" %d", ans[i]);
  printf("\n");
  return;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  eNum = 0;
  memset(head, 0, sizeof(head));
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    add(u, v);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &u);
    if (man[u][0] < 10) {
      man[u][0]++;
      man[u][man[u][0]] = i;
    }
  }
  dfs(1, 0, 1);
  initialize();
  for (int i = 1; i <= q; i++) {
    scanf("%d %d %d", &u, &v, &aa);
    lca(u, v, aa);
  }
}